JEE MainMathematicsDefinite IntegrationMCQ+4 / −1Let f(x)+2f(1x)=x2+5f(x)+2 f\left(\frac{1}{x}\right)=x^2+5f(x)+2f(x1)=x2+5 and 2g(x)−3g(12)=x,x>02 g(x)-3 g\left(\frac{1}{2}\right)=x, x\gt 02g(x)−3g(21)=x,x>0. If α=∫12f(x)dx\alpha=\int_1^2 f(x) \mathrm{d} xα=∫12f(x)dx, and β=∫12g(x)dx\beta=\int_1^2 g(x) \mathrm{d} xβ=∫12g(x)dx, then the value of 9α+β9 \alpha+\beta9α+β is :A0B10C1D11Attempt in chatSharePreviousNextMore from Definite IntegrationThe integral ∫0π1+3cos2x(x+3)sinxdx is equal to2025 · MCQThe integral −1∫23(∣π2xsin(πx))dx is equal to:2025 · MCQLet f(x) be a positive function and I1=−21∫12xf(2x(1−2x))dx and I2=−1∫2f(x(1−x))dx. Then the value of I1I2 is equal to 2025 · MCQLet for f(x)=7tan8x+7tan6x−3tan4x−3tan2x,I1=∫0π/4f(x)dx and I2=∫0π/4xf(x)dx. Then 7I1+12I2 is equal to :2025 · MCQThe value of ∫e2e4x1(e((logex)2+1)−1+e((6−logex)2+1)−1e((logex)2+1)−1)dx is2025 · MCQIf I=∫02πsin23x+cos23xsin23x dx, then ∫02Isin4x+cos4xxsinxcosx dx equals :2025 · MCQIf I(m,n)=∫01xm−1(1−x)n−1dx,m,n>0, then I(9,14)+I(10,13) is2025 · MCQIf ∫−2π2π(1+ex)96x2cos2xdx=π(απ2+β),α,β∈Z, then (α+β)2 equals2025 · MCQ